CAIE P1 2024 November — Question 4 4 marks

Exam BoardCAIE
ModuleP1 (Pure Mathematics 1)
Year2024
SessionNovember
Marks4
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Mark schemeDownload PDF ↗
TopicQuadratic trigonometric equations
TypeQuartic in sin or cos substitution
DifficultyModerate -0.3 This is a straightforward quadratic-in-disguise problem requiring substitution of u = sin²θ, solving the resulting quadratic, then finding angles. It's slightly easier than average because the technique is standard and well-practiced, though students must remember to reject the negative solution for sin²θ and find all angles in the given range.
Spec1.05j Trigonometric identities: tan=sin/cos and sin^2+cos^2=11.05o Trigonometric equations: solve in given intervals
Solve the equation \(4\sin^4\theta + 12\sin^2\theta - 7 = 0\) for \(0° \leqslant \theta \leqslant 360°\). [4]
Question 4:
AnswerMarks Guidance
4Let x=sin2
(2x+7)(2x−1)=0 or ( 2sin2+7 )( 2sin2−1 )M1 Or equivalent method.
1 1
⇒ sin2= ⇒ sin=
AnswerMarks Guidance
2 2M1 Findingsin2 and thensin(may be implied).
=45, 135, 225, 315A1 A1 A1 for any two correct values.
A1 for all correct and no others within the range.
For answers in radians, A1 only for all 4 angles.
If no (correct) working, then SC B1 for all 4 solutions.
4
AnswerMarks Guidance
QuestionAnswer Marks 
Question 4:
4 | Let x=sin2
(2x+7)(2x−1)=0 or ( 2sin2+7 )( 2sin2−1 ) | M1 | Or equivalent method.
1 1
⇒ sin2= ⇒ sin=
2 2 | M1 | Findingsin2 and thensin(may be implied).
=45, 135, 225, 315 | A1 A1 | A1 for any two correct values.
A1 for all correct and no others within the range.
For answers in radians, A1 only for all 4 angles.
If no (correct) working, then SC B1 for all 4 solutions.
4
Question | Answer | Marks | Guidance
Solve the equation $4\sin^4\theta + 12\sin^2\theta - 7 = 0$ for $0° \leqslant \theta \leqslant 360°$. [4]

\hfill \mbox{\textit{CAIE P1 2024 Q4 [4]}}
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